Question 6 - A-Level Maths

WJEC Further Unit 4 2022 June — Question 6 6 marks

Exam Board	WJEC
Module	Further Unit 4 (Further Unit 4)
Year	2022
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Standard trigonometric equations
Type	Factorization method
Difficulty	Challenging +1.8 This is a Further Maths trigonometric equation requiring multiple identities (double angle formulas, factor formulas or product-to-sum), algebraic manipulation to consolidate terms, and solving over a restricted domain. It demands more sophistication than standard A-level trig equations but follows recognizable patterns once the correct identity approach is identified.
Spec	1.05l Double angle formulae: and compound angle formulae 1.05o Trigonometric equations: solve in given intervals

Solve the equation $$\cos 2\theta - \cos 4\theta = \sin 3\theta \quad \text{for} \quad 0 \leq \theta \leq \pi$$ [6]

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Answer	Marks	Guidance
$\cos 2\theta - \cos 4\theta = -2\sin\frac{2\theta + 4\theta}{2}\sin\frac{2\theta - 4\theta}{2}$	M1	M0 no working
$-2\sin 3\theta \sin(-\theta) = \sin 3\theta$	A1
$2\sin 3\theta \sin\theta - \sin 3\theta = 0$	A1
$\sin 3\theta(2\sin\theta - 1) = 0$	A1
$\sin 3\theta = 0$	A1	FT one slip for A1A1A1. Both solutions
$\sin\theta = \frac{1}{2}$
$3\theta = 0, \pi, 2\pi, 3\pi$	A1A1	A1 each set of solutions. If A1A1, penalise -1 for use of degrees
$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi$
$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$

$\cos 2\theta - \cos 4\theta = -2\sin\frac{2\theta + 4\theta}{2}\sin\frac{2\theta - 4\theta}{2}$ | M1 | M0 no working

$-2\sin 3\theta \sin(-\theta) = \sin 3\theta$ | A1 | 

$2\sin 3\theta \sin\theta - \sin 3\theta = 0$ | A1 | 

$\sin 3\theta(2\sin\theta - 1) = 0$ | A1 | 

$\sin 3\theta = 0$ | A1 | FT one slip for A1A1A1. Both solutions
$\sin\theta = \frac{1}{2}$ | | 

$3\theta = 0, \pi, 2\pi, 3\pi$ | A1A1 | A1 each set of solutions. If A1A1, penalise -1 for use of degrees

$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi$ | | 
$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ | | 

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Solve the equation
$$\cos 2\theta - \cos 4\theta = \sin 3\theta \quad \text{for} \quad 0 \leq \theta \leq \pi$$ [6]

\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q6 [6]}}

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Q1 8 Q2 4 Q3 9 Q4 5 Q5 5 Q6 6 Q7 8 Q8 6 Q9 12 Q10 9 Q11 15 Q12 12 Q13 11 Q14 10

Answer	Marks	Guidance
\(\cos 2\theta - \cos 4\theta = -2\sin\frac{2\theta + 4\theta}{2}\sin\frac{2\theta - 4\theta}{2}\)	M1	M0 no working
\(-2\sin 3\theta \sin(-\theta) = \sin 3\theta\)	A1
\(2\sin 3\theta \sin\theta - \sin 3\theta = 0\)	A1
\(\sin 3\theta(2\sin\theta - 1) = 0\)	A1
\(\sin 3\theta = 0\)	A1	FT one slip for A1A1A1. Both solutions
\(\sin\theta = \frac{1}{2}\)
\(3\theta = 0, \pi, 2\pi, 3\pi\)	A1A1	A1 each set of solutions. If A1A1, penalise -1 for use of degrees
\(\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi\)
\(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)